Algebra Revision Sheet Questions 2 and 3 of Paper 1


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1 Algebra Revision Sheet Questions and of Paper Simple Equations Step Get rid of brackets or fractions Step Take the x s to one side of the equals sign and the numbers to the other (remember to change the sign when crossing the = ) Step Divide across by the number next to the x. Example Solve for x, (x  ) = 4x 6x = 4x Get rid of brackets by multiplying 6x 4x = + x = x Divide across by Example Solve for x, x 7 x 6 6(x  7) = (x + ) Cross multiply to get rid of the fractions 6x 4 = x + 6 Get rid of brackets by multiplying 6x x = x = x 4 Divide across by 4 Substitution Write out the question again substituting numbers for letters. Example Find the value of x 5xy when x = and y =  x 5xy Write out expression = () 5()( ) Substitute in numbers for x and y = Evaluate = 9
2 Simultaneous Equations equations we use to find values for x and y One equation will be linear (just x and y parts) One equation will be quadratic ( x or y parts) Example x + y = Equation x y 4 Equation x + y = Rearrange Equation to get x alone x = y by bringing y over the other side. x y 4 Equation ( y ) y 4 Insert the above value of x into Equation ( y) y( y)  y = 4 9 6y 6y + 4y  y = 4 Open up ( y) Remove brackets by multiplying y y 5 0 to get a quadratic equation y 4y 5 0 Divide across by ( y 5)( y ) 0 Factorise (put into brackets) y 5 = 0 and y + = 0 Let each bracket equal zero y = 5 and y =  y s to one side, numbers to the other We must now use these two values of y to get two x values. We do this by substituting the y values into Equation. x + y = Equation y = 5 then x + (5) = Substitute y = 5 into Equation x + 0 = Remove bracket by multiplying x = 0 x = 7 Evaluate So x = 7 and y = 5 Answer (7, 5) Write them as a pair, x value first y =  then x + () = Substitute y = 5 into Equation x = Remove bracket by multiplying x = + x = 5 Evaluate So x = 5 and y =  Answer (5, ) Write them as a pair, x value first
3 Algebraic Fractions These questions involve solving equations that have fractions and require you to get rid of the fractions before you can solve the equation. Step Find a lowest common denominator (usually all the bottom terms multiplied) Step Multiply each top term by any terms NOT underneath it Step Remove the common denominator Step 4 Bring all the terms to the left of the = and simplify Step 5 Solve the quadratic equation by factorising (if possible) and letting each bracket equal 0. If it cannot be factorised you must use the formula (see next page). Example Solve x x Common denominator is (x+)(x) x x ( x )() ( x )() ( x )( x ) Top terms times all terms NOT below them ( x )( x ) (x ) + (x +) = {(x +)(x )} Remove denominator (bottom section) x + x + = {x(x+) +(x )} Multiply out, open (x +)(x ) x + x + = { x x x } x + x + = { x x } x  = x x 9 x x 9 x = 0 Everything to one side. x x 8 = 0 If the equation can be factorised do so and let each bracket equal 0. If not use the FORMULA. Turn over to see how the FORMULA can be used to find our x values. The formula MUST be learned off by heart.
4 In this case we need to use the FORMULA b b 4ac x = a This formula will give us the two roots (x values) for any quadratic equation. Where possible however it is easier to factorise and let each bracket equal 0. This formula can be used in any question on the two papers to solve a quadratic and not just Q or Q. x x 8 = 0 A quadratic that cannot be factorised. a b c 8 The values of a, b and c for the formula. b ( ) b 4ac a ( ) ( ) 4( )(8) x = and x =. 7. x = and x = x = . and x =. The formula Substitute in the a, b and c values. Remove square root Split into the + and parts The roots of the equation. 4
5 Indices With questions involving indices we must break down the numbers on each side of the = to the same base number. We can then let the indices (powers) of each equal each other and solve the simple equation. To break them down we need to learn the laws of indices below (a). (b) (c) ( (e) 9 9 (f) 64 = 64 4 (g) 8 = 8 = (h) and 5 4 Example Solve for x in the equation x x x x ( 5 ) Change 5 into ) (d) x x Remove bracket to leave both sides in base 5 x = 6x Let the indices (powers) equal each other x + x = 6 x = 6 6 x Divide across by Example 4 8 ) 4 ( Solve for x in the equation = 5x 4 = The Equation 5x ( ) = Turn everything into base ( )( ) 5x Multiply the powers 5x = ( )( ) 5 x Let the powers equal each other 5 x Remove mixed fraction = (5 x) Multiply across by = 0 x Multiply to remove bracket x = 0 x = 7 7 x Divide across by 5x 5 5
6 Surds Surds are irrational numbers in the form Some important points: ab = a. b and a. b = ab Therefore 4 can be broken down into 4. 6 = 6 a a 6 6 therefore b b Terms with the same surd part can be added and subtracted Therefore = 4 6 Any surd squared is equal to the term under the root sign Therefore a 5 a 5 x x x x Equations involving surds can be solved by squaring both sides of the =. This gets rid of the surd part to leave you with a simple or quadratic equation. Example Solve 4 x 8 x 8 4 Take 4 to the other side of = x 4 4 x Square both side to remove surd part x = 6 x = 6 + x = 9 9 x = = 9 Divide across by Example ( x )( x ) x x x( x ) ( x ) x x x Open up the first bracket x x x. x ) ) x x x x Remove brackets by multiplying 9 x x Multiply 9 x x 6
7 Functions Questions with functions involve replacing the x in an expression with a number. They are generally a part (c) and can be asked in many different ways. Quite often it will ask you to put two numbers in for x and leaves you with a simultaneous equation. For example if it asks you to find f() you put in for x. If it asks you to find f(5) you put 5 into the equation for x. Below are examples showing two different types of question Example f(x) = (x + p)(x p) given that f() = 0 find the value of p. f(x) = (x + p)(x p) write out the function f() = (() + p)( p) = 0 replace x in the function with and let it = 0 (4 + p)(  p) = 0 let each bracket = p = 0 and  p = 0 p = 4 and p = Example f(x) = ax bx 8 If f() = 9 and f() = find the value of a and b f(x) = ax bx 8 write out function f() = a () b() 8 = 9 put in x = and let function = 9 a + b 8 = 9 a + b = a + b =  This will be Equation of simultaneous f(x) = ax bx 8 write out function f() = a ( ) b( ) 8 = put in x =  and let function = a b 8 = a b = + 8 a b = This will be Equation of simultaneous a + b =  Equation Simultaneous Equation a b = Equation Simultaneous Equation a = 0 b s cancel, a + a = a,  + = 0 a = 5 divide across by a + b =  Equation 5 + b =  Put a value in to get the b value b =  5 b s to one side, numbers to the other b = 6 evaluate a = 5 and b = 6 values for a and b 7
8 Factor Theorem Long Division In this type of question we are generally dealing with a cubic equation such as x x 4x = 0 There are a number of things they can ask in this question. The bold writing indicates the different things we may have to attempt to solve the equation. Show that something is a factor of the equation e.g. show that x is a factor. If x is a factor then x = is a root and by putting into the equation it will equal 0 Let us see if x is a factor by letting x = x x 4x = 0 Write out equation () 4() 0 Substitute in x = 8 + (4) 8 = = 0 This is true so x is a factor If we are NOT told the factor we must find it out through trial and error. We start off by putting x = into the equation and seeing if it equals 0 If so then x is a factor. If not we try x =  and see if this equals 0. If so then x + is a factor. If not we try x = and see if this equals 0. If so then x is a factor. If not we try x =  and so on until we have one of the factors. When we have a factor they may ask us to find OTHER factors. We do this through long division. In the above example we have found out that x is a factor of x x 4x To find the other factors divide x into x x 4x x 5x x x x 4x Divide x by x x x Multiply (x ) by 5x 4x Subtract (change signs) and divide 5x by x 5x 0x Multiply (x ) by 5x 6x Subtract (change signs) and divide 6x by x 6x Multiply (x ) by 6 0 Subtract (change signs) (x )( x 5x ) = 0 The factors of the equation (x )(x + )(x + ) = 0 We can factorise the second bracket further When we have the factors (brackets) we find the roots by letting each bracket = 0. Finding the roots means the same as solving the equation. Basically means get the x values. (x )(x + )(x + ) = 0 Factors x = 0 x + = 0 x + = 0 Let each bracket equal 0 x = x =  x =  The roots of the equation. Remember if it the question asks us to solve then we must find what x = x 8
9 Inequalities These are similar to simple equations but use the greater than, less than signs, greater than or equal to and less than or equal to signs ( and ). The rules for solving are similar to solving normal equations however one important difference is that if we decide to change all the signs we MUST change the direction of the inequality also. Example if x 4 then x 4 Change the signs, change the direction of inequality If asked to draw a number line we must pay close attention to whether the numbers are: x R  these are rational numbers, fractions, decimals etc and are illustrated on the number line with a shaded line. x Z  these are integers, all positive and negative whole numbers and are illustrated on the number line with dots. x N  there are natural numbers, positive whole numbers and are illustrated on the number line with dots. Example Solve 5x 4x for x R and illustrate on number line. 5x 4x 5x 4x x Evaluate To show this on a number line x R If the question had stated that x N or x Z we would use the following number line x N, x Z Occasionally we will be asked to find solutions for two sets in which case we need to know the meaning of the following symbols.  this means what numbers would be included in both sets. \  this means what is included in one set without the other for example A\B means what numbers are in set A but not in set B Example A is the set x 4 x Z x B is the set 5 x Z A B x 4 + x 5 A is everything less than or equal to x x 0 B is everything greater than or equal to  x 9 x  x A B therefore  x ,,,0,, (all the numbers between and inclusive, these numbers are in both set A and set B) A\B therefore x 4 (all the numbers less than 4 are in set A but are NOT in set B, they are less than but not greater than ) 9
10 Rearrange This involves using our algebra skills to rearrange equations Step Remove brackets or fractions if necessary Step Take anything with the letter we are looking for to the left of the = and everything else to the right of the =. Step If there is more than one term now on the left, factorise to get the letter alone. Step 4 Divide across by the term next to the letter we want to isolate. Example Express x in terms of a,b and c (this means get x by itself on left of the = ) ax + b = c ax = c b Bring everything with an x to the left, everything else to the right. c b x Divide across by a to isolate x a Example Express b in terms of a and c (this means get b by itself on left of the = ) 8a 5b c b 8a 5b = bc Get rid of fraction by multiplying across by b 5b bc =  8a Bring everything with a b to the left, everything else 5b + bc = 8a to the right. Change all the signs. b(5 + c) = 8a Factorise by taking out b 8a b (5 c) Divide across by (5 + c) to isolate b Example Express t in terms of p and q (this means get t by itself on left of the = ) q t p t t(p) = q  t Get rid of fraction by multiplying across by t tp + t = q Bring everything with a t to the left, everything else to the right. t(p + ) = q Factorise by taking out t q t Divide across by (p + ) to isolate t ( p ) 0
Mathematics Higher Tier, Algebraic Fractions
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